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The game works like this,

Suppose, there is a total of 900 numbers(1.00 to 9.99) from you can choose any one number to bet.

Example:

1) If you choose 1.10 number and place a bet of 100$ on it. If the generated random number is 1.10 or greater than that then you will receive 110 in return(100 * 1.10).

2) If you place a 100$ bet on number 5.00 and again generated random number is 5.00 or greater then that then you will receive 500 (100 * 5.00) in reward.

But the twist here is there is less probability for a higher number to generate a then lower number,

1) The probability of the generated number is between 1.00 - 1.49 is 60%

2) For numbers between 1.50 - 1.99. There is a probability of 30%.

3) For rest 2.00 - 9.99, there is a probability of 10%.

You can refer to this link for getting a better understanding of the game: Moon-Game

My question is if I bet on any particular number then how can I calculate my win probability?

Your answers appreciated. Thanks in advance.

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  • $\begingroup$ Are all numbers within your brackets equally likely to be chosen? $\endgroup$
    – Stephan Kolassa
    Jul 25, 2019 at 12:52
  • $\begingroup$ yes, if the chosen number is 9.99 then reward will be bet * 9.99 incase you win $\endgroup$
    – Meet Siraja
    Jul 25, 2019 at 12:53
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    $\begingroup$ That is not what I asked. My question is: are 1.11 and 1.34 equally likely (namely, with a chance of $0.60\times \frac{1}{50}$? $\endgroup$
    – Stephan Kolassa
    Jul 25, 2019 at 12:54
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    $\begingroup$ Why is the information about payoffs relevant when the question is only about probabilities? Are you sure you have formulated the question you intended to ask? $\endgroup$
    – whuber ♦
    Jul 25, 2019 at 12:55
  • $\begingroup$ Yes stephan, both numbers are equally likely to be chosen. If payout info is not relevant you can ignore it whuber. I just thought that it will be helpful for better understanding. $\endgroup$
    – Meet Siraja
    Jul 25, 2019 at 13:07

1 Answer 1

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I'm taking this question to mean that if you pick a number at random, what is the probability you'll win?

This is simply a weighted average. There's a 5/90 chance the number is chosen from bracket 1, 5/90 for bracket 2, and 8/9 for the last bracket. Multiply each case by it's winning probability and you have your answer. So,

(5/90)X(6/10) + (5/90)X(3/10) + (8/9)X(1/10) = 0.1389.

This is how the game creators make their money, as with enough people betting, only around 14% of them will win. The rest of the money goes to them.

Hope this helps!

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  • $\begingroup$ How did you calculate the chance for each bracket like 5/90, 5/90 and 8/9 $\endgroup$
    – Meet Siraja
    Jul 25, 2019 at 13:31
  • $\begingroup$ 50 of the 900 numbers are between 1.00 and 1.49. 50/900 simplifies to 5/90. Likewise for 1.50 to 1.99. The remaining 800 out of 900 numbers is 800/900 which simplifies to 8/9. $\endgroup$
    – GridAlien
    Jul 25, 2019 at 13:33
  • $\begingroup$ Okay, I get you. $\endgroup$
    – Meet Siraja
    Jul 25, 2019 at 13:34

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